# Elevator Power.

A 550 kg elevator starts from rest. It moves upward for 5.00 s with constant acceleration until it reaches its cruising speed, 1.75 m/s.
(a) What is the average power of the elevator motor during this period?
W
(b) How does this power compare with the motor power when the elevator moves at its cruising speed?
Pcruising = W

The formula for average power is avg. Power = work/change in time
However, I am not sure how to find the work in the problem since no force or angle is given...Is gravity the only force doing work and is the angle 90?

 Capuchin Posts: 5,319, Reputation: 3601 Uber Member #2 Feb 20, 2007, 09:35 AM
The angle is 0 (you don't need to worry about it, because it's a 1d problem), the motor is doing work against the force of gravity.
 tnhoots Posts: 28, Reputation: 1 New Member #3 Feb 20, 2007, 02:12 PM
Ok so the F in the problem is 9.8m/s^2. I would use cos0. However, the change in distance is unknown. Using the formula d=time * speed, I found it to be 8.75m. So, now would I use the work formula with the numbers (9.8m/s^2)(8.75m)(cos0). Which equals 85.57J. After I find that I would divide work by time (5secs) to find average power which is 17.15 W...just my thinking?
 Capuchin Posts: 5,319, Reputation: 3601 Uber Member #4 Feb 20, 2007, 02:33 PM
F = ma

Your distance is wrong because it's accelerating, you need to use suvat equations
 tnhoots Posts: 28, Reputation: 1 New Member #5 Feb 20, 2007, 03:01 PM
The force in this problem would be the mass of the elevator time the acceleration which is 1.75m/s correct. So f in the problem would be 962.5N. For distance I mult. The speed times the time. That is the only way I know to find distance.
 Capuchin Posts: 5,319, Reputation: 3601 Uber Member #6 Feb 20, 2007, 03:13 PM
Acceleration is the acceleration of gravity!

You need to use s = ut + 1/2at^2 to find distance traveled, because it is under constant acceleration.

I have to hit the sack now, but I will carry on tomorrow.
 tnhoots Posts: 28, Reputation: 1 New Member #7 Feb 20, 2007, 04:22 PM
One more try:
F=mass*acceleration
F=550*9.8=5390N
Change in distance=122.5m
Cosign of angle 0degrees

W=f*change in d*cosign0
W=5390*122.5*cosign0
? Correct?
 Capuchin Posts: 5,319, Reputation: 3601 Uber Member #8 Feb 20, 2007, 11:31 PM
Oh no, your acceleration for s = ut + 1/2 at^2 is 1.75/5